Thanks!

Spoiler:

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- Jan 13th 2011, 01:46 AMDrexel28
- Jan 13th 2011, 11:01 AMRandom Variable
My solution only involves and . Maybe your solution can be simplified.

EDIT: I found an error in my solution. Your solution is most likely correct. - Jan 13th 2011, 06:06 PMRandom Variable
I would never have thought to break up the summation like that.

- Jan 14th 2011, 07:50 AMPaulRS
Here's a nice series:

Compute: - Jan 15th 2011, 12:08 PMRandom Variable
- Jan 15th 2011, 12:16 PMchiph588@
- Jan 15th 2011, 10:20 PMDrexel28
I'm not quite sure how the last step (technically second to last, where you conclude that the sum but with is replaced with tends to zero) is achieved in your proof? Your answer is correct though. I'll leave it for someone else to do, unless no one else wants to?

- Jan 15th 2011, 11:31 PMchiph588@
- Jan 16th 2011, 04:11 AMPaulRS
- Jan 16th 2011, 01:54 PMchiph588@
- Jan 16th 2011, 03:48 PMchiph588@
Compute

- Jan 16th 2011, 09:00 PMDrexel28
- Jan 16th 2011, 09:04 PMDrexel28
What about ?

- Jan 16th 2011, 09:30 PMchiph588@
- Jan 16th 2011, 09:33 PMDrexel28